Division by a one-digit number: place-value methods

Learning goals

Divides multiples of ten using related division facts.
Splits a dividend into parts divisible by the divisor.
Combines partial quotients to find the full quotient.
Explains regrouping between hundreds, tens and ones in division.
Predicts how changing the dividend or divisor changes the quotient.
Estimates a quotient using compatible numbers.

Difficulty: 1 – 5

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Division by a one-digit number: place-value methods 1

A school has $600$ counting blocks. It shares them equally among $3$ classes. Think of the blocks as hundreds. How many blocks does each class get?

Hints

- How many hundreds make $600$? - Share the hundreds first, using a familiar division fact. - Turn the number of hundreds back into blocks.

Solution

1. $600$ is $6$ hundreds. 2. Share $6$ hundreds among $3$ classes: each gets $2$ hundreds. 3. So each class gets $200$ blocks.

Answer

$200$ blocks per class.

Division by a one-digit number: place-value methods 2

Split $96$ into $80+16$ to calculate $96\div4$. What does each part give, and what is the total quotient?

Hints

- Imagine sharing $96$ counters equally into four groups. Does each part of the split belong to all four groups? - Think about how many groups of four are in $80$, then in $16$. - Check the result by multiplying the quotient by four.

Solution

1. $80\div4=20$. 2. $16\div4=4$. 3. Add the quotients: $20+4=24$, so $96\div4=24$.

Answer

$80\div4=20$, $16\div4=4$, and $96\div4=24$.

Division by a one-digit number: place-value methods 3

Divide $606$ equally into $3$ shares using hundreds, tens, and ones. Which place in the quotient has a $0$, and why?

Hints

- Think about each place in the starting number separately. - If there is nothing to share in a place, what will each group receive there? - Check by multiplying the three-digit quotient by $3$.

Solution

1. Each share gets $6\div3=2$ hundreds. 2. There are no tens, so each share gets $0$ tens. 3. Each share gets $6\div3=2$ ones, so the quotient is $202$.

Answer

$606\div3=202$. The tens place is $0$ because no tens need to be shared.

Division by a one-digit number: place-value methods 4

Use basic division facts and place-value patterns to find each quotient mentally. a) $360 \div 4$ b) $420 \div 6$ c) $720 \div 9$ d) $4000 \div 8$

Hints

- Look for a related basic division fact. - Use place value to account for the factors of $10$. - Check each quotient with multiplication.

Solution

1. For a), use $36 \div 4 = 9$, so $360 \div 4 = 90$. 2. For b), use $42 \div 6 = 7$, so $420 \div 6 = 70$. 3. For c), use $72 \div 9 = 8$, so $720 \div 9 = 80$. 4. For d), use $40 \div 8 = 5$, so $4000 \div 8 = 500$.

Answer

a) $90$ b) $70$ c) $80$ d) $500$

Division by a one-digit number: place-value methods 5

Find each quotient. a) $4000\div4$ b) $1200\div4$ c) $28\div4$ d) $5228\div4$

Hints

- Which numbers from a), b), and c) combine to make $5228$? - Divide each convenient part separately. - Add the partial quotients to find the final quotient.

Solution

1. $4000 \div 4 = 1000$. 2. $1200 \div 4 = 300$. 3. $28 \div 4 = 7$. 4. Since $5228 = 4000 + 1200 + 28$, add the partial quotients: $1000 + 300 + 7 = 1307$.

Answer

a) $1000$ b) $300$ c) $7$ d) $1307$

Division by a one-digit number: place-value methods 6

Find each quotient. a) $52\div4$ b) $78\div3$ c) $96\div6$ d) $85\div5$

Hints

- Choose a nearby multiple of the divisor. - Break the dividend into that multiple and a remaining part. - Divide both parts by the divisor. - Add the two partial quotients.

Solution

1. For part a, break $52$ into $40 + 12$. Then $40 \div 4 = 10$ and $12 \div 4 = 3$, so $10 + 3 = 13$. 2. For part b, break $78$ into $60 + 18$. Then $60 \div 3 = 20$ and $18 \div 3 = 6$, so $20 + 6 = 26$. 3. For part c, break $96$ into $60 + 36$. Then $60 \div 6 = 10$ and $36 \div 6 = 6$, so $10 + 6 = 16$. 4. For part d, break $85$ into $50 + 35$. Then $50 \div 5 = 10$ and $35 \div 5 = 7$, so $10 + 7 = 17$.

Answer

a) $13$ b) $26$ c) $16$ d) $17$

Division by a one-digit number: place-value methods 7

How many groups of $4$ are in $92$? How many groups of $3$ are in $75$?

Hints

- Decompose each dividend into numbers that are easy to divide. - Use nearby multiples of the divisor. - Add the partial quotients.

Solution

1. Decompose $92$ as $80 + 12$: $80 \div 4 = 20$ and $12 \div 4 = 3$. Thus $92 \div 4 = 23$. 2. Decompose $75$ as $60 + 15$: $60 \div 3 = 20$ and $15 \div 3 = 5$. Thus $75 \div 3 = 25$.

Answer

There are $23$ groups of $4$ in $92$, and $25$ groups of $3$ in $75$.

Division by a one-digit number: place-value methods 8

Find each quotient. a) $168\div3$ b) $455\div7$

Hints

- Choose two parts that are both multiples of the divisor. - Divide each part separately. - Add the partial quotients.

Solution

1. For part a, break $168$ into $150 + 18$. Then $150 \div 3 = 50$ and $18 \div 3 = 6$, so $50 + 6 = 56$. 2. For part b, break $455$ into $420 + 35$. Then $420 \div 7 = 60$ and $35 \div 7 = 5$, so $60 + 5 = 65$.

Answer

a) $56$ b) $65$

Division by a one-digit number: place-value methods 9

A coach has $96$ tennis balls. The coach puts exactly $6$ balls in each basket. How many baskets can be filled completely?

Hints

- Break $96$ into two numbers that are easy to divide by $6$. - Divide each part by $6$. - Add the partial quotients.

Solution

1. Divide the total number of tennis balls by the number in each basket: $96 \div 6$. 2. Break $96$ into compatible numbers: $96 = 60 + 36$. 3. Divide each part: $60 \div 6 = 10$ and $36 \div 6 = 6$. 4. Add the partial quotients: $10 + 6 = 16$.

Answer

The coach can fill $16$ baskets completely.

Division by a one-digit number: place-value methods 10

A bakery made $144$ dinner rolls. The baker divides them equally among $6$ display baskets. How many rolls are in each basket?

Hints

- Break $144$ into two numbers that are easy to divide by $6$. - Divide each part by $6$. - Add the partial quotients.

Solution

1. Divide the total number of rolls by the number of baskets: $144 \div 6$. 2. Break $144$ into compatible numbers: $144 = 120 + 24$. 3. Divide each part: $120 \div 6 = 20$ and $24 \div 6 = 4$. 4. Add the partial quotients: $20 + 4 = 24$.

Answer

Each basket contains $24$ rolls.

Division by a one-digit number: place-value methods 11

A large building-block model of a ship has $864$ pieces. A smaller lifeboat model has one-eighth as many pieces. How many pieces are in the lifeboat model?

Hints

- What operation finds one-eighth of a number? - Break $864$ into two numbers that are easy to divide by $8$. - Divide each part and combine the results.

Solution

1. To find one-eighth of $864$, divide by $8$: $864 \div 8$. 2. Break $864$ into compatible numbers: $864 = 800 + 64$. 3. Divide each part: $800 \div 8 = 100$ and $64 \div 8 = 8$. 4. Add the partial quotients: $100 + 8 = 108$.

Answer

The lifeboat model has $108$ pieces.

Division by a one-digit number: place-value methods 12

A beekeeper pours $450\,\text{g}$ of honey equally into $5$ small jars. How many grams of honey are in each jar?

Hints

- Divide the total amount of honey equally among $5$ jars. - Use place value and a related basic division fact. - Include the unit in your answer.

Solution

1. Divide the total mass of honey by the number of jars: $450 \div 5$. 2. Use $45 \div 5 = 9$ and place-value reasoning to get $450 \div 5 = 90$. 3. Each jar contains $90\,\text{g}$ of honey.

Answer

Each jar contains $90\,\text{g}$ of honey.

Division by a one-digit number: place-value methods 13

A garden center divides $252$ seedlings equally among $6$ planters. How many seedlings go in each planter?

Hints

- Break $252$ into two multiples of $6$. - Divide both parts by $6$. - Add the partial quotients.

Solution

1. Break $252$ into $240 + 12$, two numbers divisible by $6$. 2. Divide each part: $240 \div 6 = 40$ and $12 \div 6 = 2$. 3. Add the partial quotients: $40 + 2 = 42$.

Answer

Each planter gets $42$ seedlings.

Division by a one-digit number: place-value methods 14

A professional cyclist rides $160$ miles in $4$ hours. What is the cyclist's average distance per hour?

Hints

- The distance is shared equally across $4$ hours. - Use place value and a related basic division fact. - State the distance traveled in one hour.

Solution

1. Divide the total distance by the number of hours: $160 \div 4$. 2. Use $16 \div 4 = 4$ and place-value reasoning to get $160 \div 4 = 40$. 3. The cyclist averages $40$ miles per hour.

Answer

The cyclist averages $40$ miles per hour.

Division by a one-digit number: place-value methods 15

Evaluate each quotient. a) $420 \div 6$ b) $560 \div 4$ c) $750 \div 3$

Hints

- Decompose each dividend into easy-to-divide parts. - Use related basic facts. - Check with multiplication.

Solution

1. a) Use $42 \div 6 = 7$, so $420 \div 6 = 70$. 2. b) Decompose $560$: $400 \div 4 = 100$ and $160 \div 4 = 40$. Thus $560 \div 4 = 140$. 3. c) Decompose $750$: $600 \div 3 = 200$ and $150 \div 3 = 50$. Thus $750 \div 3 = 250$.

Answer

a) $70$ b) $140$ c) $250$

Division by a one-digit number: place-value methods 16

A package of $3$ identical notepads costs $\$7.20$. How much does one notepad cost?

Hints

- Convert the total price to cents. - Divide the total cost equally among $3$ notepads. - Convert the result back to dollars.

Solution

1. Write the total price in cents: $\$7.20 = 720\text{ cents}$. 2. Divide the total price by the number of notepads: $720 \div 3$. 3. Break $720$ into compatible numbers: $600 \div 3 = 200$ and $120 \div 3 = 40$. Then add: $200 + 40 = 240$ cents. 4. Convert $240$ cents to dollars: $240\text{ cents} = \$2.40$.

Answer

One notepad costs $\$2.40$.

Division by a one-digit number: place-value methods 17

A sports club is ordering T-shirts. Each shirt costs $\$8$, and the club has saved $\$480$. How many T-shirts can the club buy?

Hints

- Divide the total amount saved by the price of one shirt. - Use place value and a related basic division fact. - Check by multiplying the number of shirts by the price of one shirt.

Solution

1. Divide the total amount of money by the cost of each shirt: $480 \div 8$. 2. Use $48 \div 8 = 6$ and place-value reasoning to get $480 \div 8 = 60$.

Answer

The club can buy $60$ T-shirts.

Division by a one-digit number: place-value methods 18

Find each quotient. Break apart the dividends to make the division easier. a) $51 \div 3$ b) $68 \div 4$ c) $95 \div 5$ d) $84 \div 6$ e) $91 \div 7$ f) $104 \div 8$

Hints

- Break each dividend into two familiar multiples of the divisor. - Dividing ten groups of the divisor is often a useful first part. - Add the two partial quotients.

Solution

1. $51 \div 3 = (30 \div 3) + (21 \div 3) = 10 + 7 = 17$. 2. $68 \div 4 = (40 \div 4) + (28 \div 4) = 10 + 7 = 17$. 3. $95 \div 5 = (50 \div 5) + (45 \div 5) = 10 + 9 = 19$. 4. $84 \div 6 = (60 \div 6) + (24 \div 6) = 10 + 4 = 14$. 5. $91 \div 7 = (70 \div 7) + (21 \div 7) = 10 + 3 = 13$. 6. $104 \div 8 = (80 \div 8) + (24 \div 8) = 10 + 3 = 13$.

Answer

a) $17$ b) $17$ c) $19$ d) $14$ e) $13$ f) $13$

Division by a one-digit number: place-value methods 19

Two divisions can be solved by splitting the number being divided into easier parts. Match each division to a split. Then divide both parts by $7$ and add the partial quotients. A) $112\div7$ B) $119\div7$ Splits: 1) $70+49$ 2) $70+42$

Hints

- Check which two parts add up to each number being divided. - Each part can be divided by the same number as the whole. - Add the two partial quotients after you have calculated them.

Solution

1. The split $70+42=112$ matches A, so A uses split 2. 2. Divide the two parts: $70\div7=10$, $42\div7=6$, and $10+6=16$. 3. The split $70+49=119$ matches B, so B uses split 1. 4. Divide the two parts: $70\div7=10$, $49\div7=7$, and $10+7=17$.

Answer

A) Split 2; $112\div7=16$. B) Split 1; $119\div7=17$.

Division by a one-digit number: place-value methods 20

A box contains $72$ chocolates. The chocolates are shared equally among $4$ children. How many chocolates does each child get?

Hints

- Break $72$ into two numbers that are easy to divide by $4$. - Divide each part by $4$. - Add the two partial quotients.

Solution

1. Divide the total number of chocolates by the number of children: $72 \div 4$. 2. Break $72$ into compatible numbers: $72 = 40 + 32$. 3. Divide each part: $40 \div 4 = 10$ and $32 \div 4 = 8$. 4. Add the partial quotients: $10 + 8 = 18$.

Answer

Each child gets $18$ chocolates.

Division by a one-digit number: place-value methods 21

A gardener has $156$ seedlings and plants them in $6$ equal rows. How many seedlings are in each row?

Hints

- Break $156$ into two multiples of $6$. - Divide each part by $6$. - Add the partial quotients.

Solution

1. Divide the total number of seedlings by the number of rows: $156 \div 6$. 2. Break $156$ into compatible numbers: $156 = 120 + 36$. 3. Divide each part: $120 \div 6 = 20$ and $36 \div 6 = 6$. 4. Add the partial quotients: $20 + 6 = 26$.

Answer

There are $26$ seedlings in each row.

Division by a one-digit number: place-value methods 22

An elementary school receives a donation of $\$744$. The money is divided equally among the school's $6$ classes. How much money does each class receive?

Hints

- Break $744$ into numbers that are easy to divide by $6$. - Divide each part by $6$. - Add the partial quotients and include the dollar sign.

Solution

1. Divide the total donation by the number of classes: $744 \div 6$. 2. Decompose $744$ into compatible numbers: $744 = 600 + 120 + 24$. 3. Divide each part: $600 \div 6 = 100$, $120 \div 6 = 20$, and $24 \div 6 = 4$. 4. Add the partial quotients: $100 + 20 + 4 = 124$.

Answer

Each class receives $\$124$.

Division by a one-digit number: place-value methods 23

A school spends $\$168$ on identical art kits. Each art kit costs $\$7$. How many art kits does the school buy?

Hints

- Divide the total cost by the price of one kit. - Break $168$ into multiples of $7$. - Check the quotient by multiplication.

Solution

1. Divide the total cost by the cost of one art kit: $168 \div 7$. 2. Break $168$ into compatible numbers: $168 = 140 + 28$. 3. Divide each part: $140 \div 7 = 20$ and $28 \div 7 = 4$. 4. Add the partial quotients: $20 + 4 = 24$.

Answer

The school buys $24$ art kits.

Division by a one-digit number: place-value methods 24

Determine the number of digits in each quotient without carrying out the full division. a) $8550\div5$ b) $2412\div6$ c) $3180\div9$

Hints

- Compare the first digit of each dividend with the divisor. - If the first digit is too small to divide by the divisor, division starts with the first two digits. - Use the starting place to decide how many quotient digits there will be; you do not need the exact quotient.

Solution

1. For a), the leading digit $8$ is at least the divisor $5$, so division begins in the thousands place. The quotient has $4$ digits. 2. For b), the leading digit $2$ is less than $6$, so division begins with $24$ in the first two places. The quotient has $3$ digits. 3. For c), the leading digit $3$ is less than $9$, so division begins with $31$ in the first two places. The quotient has $3$ digits.

Answer

a) $4$ digits b) $3$ digits c) $3$ digits

Division by a one-digit number: place-value methods 25

Three school classes earn a total of $\$255$ at a fundraiser. The money is divided equally among the three classes. How much money does each class receive?

Hints

- Break $255$ into two multiples of $3$. - Divide both parts by $3$. - Add the partial quotients.

Solution

1. Divide the total by $3$: $255 \div 3$. 2. Break $255$ into $240 + 15$. 3. Divide each part: $240 \div 3 = 80$ and $15 \div 3 = 5$. 4. Add the partial quotients: $80 + 5 = 85$.

Answer

Each class receives $\$85$.

Division by a one-digit number: place-value methods 26

A beverage company has $432$ bottles of juice. How many cases are needed if each case holds $6$ bottles? How many cases are needed if each case holds $8$ bottles instead?

Hints

- Solve two separate division problems. - Break $432$ into compatible numbers for each divisor. - Check whether the larger case size gives more or fewer cases.

Solution

1. For cases of $6$, divide: $432 \div 6$. Break apart $432 = 420 + 12$, so $420 \div 6 = 70$ and $12 \div 6 = 2$. Therefore, $70 + 2 = 72$. 2. For cases of $8$, divide: $432 \div 8$. Break apart $432 = 400 + 32$, so $400 \div 8 = 50$ and $32 \div 8 = 4$. Therefore, $50 + 4 = 54$.

Answer

The company needs $72$ cases that hold $6$ bottles or $54$ cases that hold $8$ bottles.

Division by a one-digit number: place-value methods 27

A beekeeper has $360$ jars of honey and packs them into cartons that hold $6$ jars each. a) How many cartons can the beekeeper fill? b) If the beekeeper uses larger cartons that hold $9$ jars each, will more or fewer cartons be needed? Explain without calculating the exact number of larger cartons.

Hints

- Divide $360$ by $6$ for part a. - For part b, compare the two carton capacities. - Think about how group size affects the number of groups when the total stays the same.

Solution

1. For part a, divide the number of jars by the number in each carton: $360 \div 6 = 60$. 2. For part b, each larger carton holds more jars. Packing the same total into larger groups produces fewer groups, so fewer cartons are needed.

Answer

a) The beekeeper can fill $60$ cartons. b) Fewer cartons are needed because each carton holds more jars.

Division by a one-digit number: place-value methods 28

A beekeeper harvests $480\,\text{kg}$ of honey and divides it equally among $4$ large barrels. How many kilograms of honey are in each barrel?

Hints

- Break $480$ into two numbers that are easy to divide by $4$. - Divide each part by $4$. - Add the partial quotients and include the unit.

Solution

1. Divide the total mass by the number of barrels: $480 \div 4$. 2. Break $480$ into compatible numbers: $480 = 400 + 80$. 3. Divide each part: $400 \div 4 = 100$ and $80 \div 4 = 20$. 4. Add the partial quotients: $100 + 20 = 120$.

Answer

Each barrel contains $120\,\text{kg}$ of honey.

Division by a one-digit number: place-value methods 29

Find each quotient using partial quotients. Break each dividend into two or more numbers that are easy to divide. a) $360 \div 3$ b) $520 \div 4$ c) $750 \div 5$

Hints

- Break each dividend into familiar multiples of the divisor. - Divide the parts separately, then add the partial quotients.

Solution

1. For part a, $300 \div 3 = 100$ and $60 \div 3 = 20$. Then $100 + 20 = 120$. 2. For part b, $400 \div 4 = 100$ and $120 \div 4 = 30$. Then $100 + 30 = 130$. 3. For part c, $500 \div 5 = 100$ and $250 \div 5 = 50$. Then $100 + 50 = 150$.

Answer

a) $120$ b) $130$ c) $150$

Division by a one-digit number: place-value methods 30

A toy store has $540$ marbles and packs exactly $6$ marbles in each small bag. How many bags can the store fill?

Hints

- Divide the total number of marbles by $6$. - Use place value and a related basic division fact. - Check your quotient by multiplication.

Solution

1. Divide the total number of marbles by the number in each bag: $540 \div 6$. 2. Use $54 \div 6 = 9$ and place-value reasoning to get $540 \div 6 = 90$.

Answer

The store can fill $90$ bags.

Division by a one-digit number: place-value methods 31

A school has $825$ pencils to pack for classrooms. Each pack holds $5$ pencils. a) How many full packs can the school make? b) If each pack could hold up to $10$ pencils instead, would the school need more or fewer packs? Explain without finding the exact number.

Hints

- Think about splitting all the pencils into equal groups. - Can you break the total into smaller amounts that are easier to divide? - For part b, compare how many pencils fit in one pack.

Solution

1. Divide the pencils into packs: $825 \div 5$. 2. Split $825 = 500 + 300 + 25$. Then $500 \div 5 = 100$, $300 \div 5 = 60$, and $25 \div 5 = 5$. 3. Add: $100 + 60 + 5 = 165$ full packs. 4. Packs that hold up to $10$ pencils would hold more each, so fewer packs would be needed.

Answer

a) $165$ full packs. b) Fewer packs, because each pack could hold more pencils.

Division by a one-digit number: place-value methods 32

Six children earn $\$168$ by selling toys at a yard sale. They divide the money equally. How much money does each child receive?

Hints

- Break $168$ into two multiples of $6$. - Divide each part by $6$. - Add the partial quotients.

Solution

1. Break $168$ into $120 + 48$, two numbers divisible by $6$. 2. Divide each part: $120 \div 6 = 20$ and $48 \div 6 = 8$. 3. Add the partial quotients: $20 + 8 = 28$.

Answer

Each child receives $\$28$.

Division by a one-digit number: place-value methods 33

A gardener has $161$ flower bulbs and plants exactly $7$ bulbs in each row. How many complete rows can the gardener plant?

Hints

- Break $161$ into two multiples of $7$. - Divide each part by $7$. - Add the partial quotients.

Solution

1. Divide the total number of bulbs by the number in each row: $161 \div 7$. 2. Break $161$ into compatible numbers: $161 = 140 + 21$. 3. Divide each part: $140 \div 7 = 20$ and $21 \div 7 = 3$. 4. Add the partial quotients: $20 + 3 = 23$.

Answer

The gardener can plant $23$ complete rows.

Division by a one-digit number: place-value methods 34

Break each dividend into two useful parts and fill in the blanks. a) $72 \div 3 = (60 \div 3) + (\square \div 3) = \square + \square = \square$ b) $95 \div 5 = (50 \div 5) + (\square \div 5) = \square + \square = \square$ c) $84 \div 4 = (80 \div 4) + (\square \div 4) = \square + \square = \square$

Hints

- Choose two parts that are both multiples of the divisor. - Find the remaining part after choosing the first partial dividend. - Add the partial quotients.

Solution

1. For part a, $72 = 60 + 12$. Then $(60 \div 3) + (12 \div 3) = 20 + 4 = 24$. 2. For part b, $95 = 50 + 45$. Then $(50 \div 5) + (45 \div 5) = 10 + 9 = 19$. 3. For part c, $84 = 80 + 4$. Then $(80 \div 4) + (4 \div 4) = 20 + 1 = 21$.

Answer

a) $72 \div 3 = (60 \div 3) + (12 \div 3) = 20 + 4 = 24$ b) $95 \div 5 = (50 \div 5) + (45 \div 5) = 10 + 9 = 19$ c) $84 \div 4 = (80 \div 4) + (4 \div 4) = 20 + 1 = 21$

Division by a one-digit number: place-value methods 35

Evaluate each quotient. a) $480 \div 2$ b) $606 \div 3$ c) $909 \div 9$ d) $840 \div 4$ e) $550 \div 5$

Hints

- Decompose each dividend into place-value parts that are easy to divide. - Divide each part by the divisor. - Add the partial quotients.

Solution

1. a) Decompose $480$ as $400 + 80$: $400 \div 2 = 200$ and $80 \div 2 = 40$. The quotient is $240$. 2. b) Decompose $606$ as $600 + 6$: $600 \div 3 = 200$ and $6 \div 3 = 2$. The quotient is $202$. 3. c) Decompose $909$ as $900 + 9$: $900 \div 9 = 100$ and $9 \div 9 = 1$. The quotient is $101$. 4. d) Decompose $840$ as $800 + 40$: $800 \div 4 = 200$ and $40 \div 4 = 10$. The quotient is $210$. 5. e) Decompose $550$ as $500 + 50$: $500 \div 5 = 100$ and $50 \div 5 = 10$. The quotient is $110$.

Answer

a) $240$ b) $202$ c) $101$ d) $210$ e) $110$

Division by a one-digit number: place-value methods 36

Find each quotient. a) $428\div4$ b) $618\div6$ c) $927\div9$ d) $515\div5$

Hints

- Break each dividend into a large place-value number and a smaller multiple of the divisor. - Divide both parts. - Add the partial quotients.

Solution

1. For part a, $428 = 400 + 28$. Then $400 \div 4 = 100$ and $28 \div 4 = 7$, so the quotient is $107$. 2. For part b, $618 = 600 + 18$. Then $600 \div 6 = 100$ and $18 \div 6 = 3$, so the quotient is $103$. 3. For part c, $927 = 900 + 27$. Then $900 \div 9 = 100$ and $27 \div 9 = 3$, so the quotient is $103$. 4. For part d, $515 = 500 + 15$. Then $500 \div 5 = 100$ and $15 \div 5 = 3$, so the quotient is $103$.

Answer

a) $107$ b) $103$ c) $103$ d) $103$

Division by a one-digit number: place-value methods 37

Find each quotient. a) $378\div6$ b) $532\div7$

Hints

- Find a large familiar multiple of the divisor below the dividend. - Divide that part and the remaining part separately. - Add the partial quotients.

Solution

1. For part a, break $378$ into $360 + 18$. Then $360 \div 6 = 60$ and $18 \div 6 = 3$, so $60 + 3 = 63$. 2. For part b, break $532$ into $490 + 42$. Then $490 \div 7 = 70$ and $42 \div 7 = 6$, so $70 + 6 = 76$.

Answer

a) $378 \div 6 = 63$ b) $532 \div 7 = 76$

Division by a one-digit number: place-value methods 38

A delivery truck brings $544$ boxes to a supermarket. A worker divides all the boxes equally among $8$ rolling carts. How many boxes are placed on each cart? Show your work.

Hints

- Break $544$ into two multiples of $8$. - Divide each part by $8$. - Add the partial quotients.

Solution

1. Divide the total number of boxes by the number of carts: $544 \div 8$. 2. Break $544$ into compatible numbers: $544 = 480 + 64$. 3. Divide each part: $480 \div 8 = 60$ and $64 \div 8 = 8$. 4. Add the partial quotients: $60 + 8 = 68$.

Answer

Each rolling cart holds $68$ boxes.

Division by a one-digit number: place-value methods 39

Find the quotients. a) $640 \div 2$ b) $640 \div 4$ c) $640 \div 8$ What rule describes the quotients when the divisor doubles each time?

Hints

- Compare the divisors $2$, $4$, and $8$. - Think about what happens to a quotient when the divisor becomes larger. - Use the first quotient to find the second, then the third. - Describe the inverse change between divisor and quotient.

Solution

1. $640 \div 2 = 320$. 2. The divisor doubles from $2$ to $4$, so the quotient is halved: $640 \div 4 = 160$. 3. The divisor doubles again from $4$ to $8$, so the quotient is halved again: $640 \div 8 = 80$. 4. With a fixed dividend, doubling the divisor halves the quotient.

Answer

a) $320$ b) $160$ c) $80$ When the divisor doubles, the quotient is halved.

Division by a one-digit number: place-value methods 40

Find each quotient. a) $84\div3$ b) $75\div5$ c) $92\div4$ d) $108\div6$

Hints

- Break each dividend into two multiples of the divisor. - Start with a large multiple that is easy to divide. - Divide the remaining part separately. - Add the partial quotients.

Solution

1. For part a, $84 = 60 + 24$. Then $60 \div 3 = 20$ and $24 \div 3 = 8$, so the quotient is $28$. 2. For part b, $75 = 50 + 25$. Then $50 \div 5 = 10$ and $25 \div 5 = 5$, so the quotient is $15$. 3. For part c, $92 = 80 + 12$. Then $80 \div 4 = 20$ and $12 \div 4 = 3$, so the quotient is $23$. 4. For part d, $108 = 60 + 48$. Then $60 \div 6 = 10$ and $48 \div 6 = 8$, so the quotient is $18$.

Answer

a) $28$ b) $15$ c) $23$ d) $18$

Division by a one-digit number: place-value methods 41

Complete the partial-quotients work. Problem 1: $824 \div 4$ $800 \div 4 = \square$ $24 \div 4 = \square$ Quotient: $\square$ Problem 2: $918 \div 9$ $900 \div 9 = \square$ $18 \div 9 = \square$ Quotient: $\square$

Hints

- Divide the large place-value part first. - Divide the remaining part by the same divisor. - Add the partial quotients.

Solution

1. For Problem 1, $800 \div 4 = 200$ and $24 \div 4 = 6$. Then $200 + 6 = 206$. 2. For Problem 2, $900 \div 9 = 100$ and $18 \div 9 = 2$. Then $100 + 2 = 102$.

Answer

Problem 1: $800 \div 4 = 200$, $24 \div 4 = 6$, quotient: $206$ Problem 2: $900 \div 9 = 100$, $18 \div 9 = 2$, quotient: $102$

Division by a one-digit number: place-value methods 42

Find each quotient. a) $147\div7$ b) $255\div5$ c) $368\div8$ d) $426\div6$

Hints

- Find a nearby multiple of the divisor with a zero in the ones place. - Break the dividend into that large part and a smaller remaining part. - Divide both parts separately. - Add the partial quotients.

Solution

1. For part a, $(140 \div 7) + (7 \div 7) = 20 + 1 = 21$. 2. For part b, $(250 \div 5) + (5 \div 5) = 50 + 1 = 51$. 3. For part c, $(320 \div 8) + (48 \div 8) = 40 + 6 = 46$. 4. For part d, $(420 \div 6) + (6 \div 6) = 70 + 1 = 71$.

Answer

a) $21$ b) $51$ c) $46$ d) $71$

Division by a one-digit number: place-value methods 43

Find each quotient. a) $424\div4$ b) $721\div7$ c) $636\div6$ d) $918\div9$

Hints

- Break each dividend into a hundreds part and a remaining part. - Divide the hundreds part by the divisor. - Divide the remaining part by the divisor. - Add the partial quotients.

Solution

1. For part a, $424 = 400 + 24$. Then $400 \div 4 = 100$ and $24 \div 4 = 6$, so the quotient is $106$. 2. For part b, $721 = 700 + 21$. Then $700 \div 7 = 100$ and $21 \div 7 = 3$, so the quotient is $103$. 3. For part c, $636 = 600 + 36$. Then $600 \div 6 = 100$ and $36 \div 6 = 6$, so the quotient is $106$. 4. For part d, $918 = 900 + 18$. Then $900 \div 9 = 100$ and $18 \div 9 = 2$, so the quotient is $102$.

Answer

a) $106$ b) $103$ c) $106$ d) $102$

Division by a one-digit number: place-value methods 44

Find the quotients for $360$. a) $360 \div 2$, $360 \div 4$, and $360 \div 8$ b) $360 \div 3$, $360 \div 6$, and $360 \div 9$

Hints

- Compare the divisors within each row. - Use a known quotient to find a related one when the divisor doubles. - In b), compare $9$ with $3$. - You may decompose $360$ into smaller numbers that are easier to divide.

Solution

1. a) $360 \div 2 = 180$. Doubling the divisor gives $360 \div 4 = 90$, and doubling it again gives $360 \div 8 = 45$. 2. b) $360 \div 3 = 120$. Doubling the divisor gives $360 \div 6 = 60$. Since $9$ is three times $3$, $360 \div 9 = 120 \div 3 = 40$.

Answer

a) $180$, $90$, $45$ b) $120$, $60$, $40$

Division by a one-digit number: place-value methods 45

Divide $840$ by each number. a) $2$ and $4$ b) $3$ and $6$ c) $7$ and $8$

Hints

- Break $840$ into parts that are easy to divide. - Think about how dividing by $4$ is related to dividing by $2$. - Think about how dividing by $6$ is related to dividing by $3$. - For part c, use multiplication facts to check each quotient.

Solution

1. $840\div2=420$, so $840\div4=210$, which is half of $420$. 2. $840\div3=280$, so $840\div6=140$, which is half of $280$. 3. $840\div7=120$, and $840\div8=105$.

Answer

a) $420$ and $210$ b) $280$ and $140$ c) $120$ and $105$

Division by a one-digit number: place-value methods 46

Find each quotient. What do you notice when you compare the results? a) $420 \div 3$ b) $560 \div 4$ c) $700 \div 5$ d) $840 \div 6$

Hints

- Break each dividend into two numbers that are easy to divide. - Divide each part, then add the partial quotients. - Compare the four results.

Solution

1. $420 \div 3 = (300 \div 3) + (120 \div 3) = 100 + 40 = 140$. 2. $560 \div 4 = (400 \div 4) + (160 \div 4) = 100 + 40 = 140$. 3. $700 \div 5 = (500 \div 5) + (200 \div 5) = 100 + 40 = 140$. 4. $840 \div 6 = (600 \div 6) + (240 \div 6) = 100 + 40 = 140$. 5. All four quotients are equal.

Answer

a) $140$ b) $140$ c) $140$ d) $140$ All four quotients are equal.

Division by a one-digit number: place-value methods 47

Find each quotient. Which result does not match the others? a) $480 \div 3$ b) $640 \div 4$ c) $800 \div 5$ d) $780 \div 6$

Hints

- Break each dividend into convenient parts. - Record every quotient before comparing them. - Check a quotient by multiplying it by the divisor.

Solution

1. $480 \div 3 = (300 \div 3) + (180 \div 3) = 100 + 60 = 160$. 2. $640 \div 4 = (400 \div 4) + (240 \div 4) = 100 + 60 = 160$. 3. $800 \div 5 = (500 \div 5) + (300 \div 5) = 100 + 60 = 160$. 4. $780 \div 6 = (600 \div 6) + (180 \div 6) = 100 + 30 = 130$. 5. Part d) is the only quotient that is not $160$.

Answer

d) does not match. $780 \div 6 = 130$, while the other three quotients are $160$.

Division by a one-digit number: place-value methods 48

Divide $160$, $320$, $480$, $640$, and $800$ by $4$. Then divide the same numbers by $8$. What do you notice when you compare each pair of quotients?

Hints

- Compare the divisors $4$ and $8$. - Consider what happens to a quotient when the divisor doubles. - Compare the two quotients for one dividend before checking the rest.

Solution

1. Dividing by $4$ gives $40$, $80$, $120$, $160$, and $200$. 2. Dividing by $8$ gives $20$, $40$, $60$, $80$, and $100$. 3. Since $8$ is twice $4$, dividing the same number by $8$ gives half the quotient obtained by dividing it by $4$.

Answer

Quotients with $4$: $40$, $80$, $120$, $160$, $200$ Quotients with $8$: $20$, $40$, $60$, $80$, $100$ Each quotient with $8$ is half the matching quotient with $4$.

Division by a one-digit number: place-value methods 49

Find each quotient. Use a related basic fact and place value. a) $320 \div 4$ b) $450 \div 9$ c) $630 \div 7$ d) $210 \div 3$ e) $810 \div 9$

Hints

- Identify the related basic division fact. - Think of the dividend as a number of tens. - Use a multiplication fact to check each quotient.

Solution

1. Since $32 \div 4 = 8$, $320 \div 4 = 80$. 2. Since $45 \div 9 = 5$, $450 \div 9 = 50$. 3. Since $63 \div 7 = 9$, $630 \div 7 = 90$. 4. Since $21 \div 3 = 7$, $210 \div 3 = 70$. 5. Since $81 \div 9 = 9$, $810 \div 9 = 90$.

Answer

a) $80$ b) $50$ c) $90$ d) $70$ e) $90$

Division by a one-digit number: place-value methods 50

Find each quotient. a) $639\div3$ b) $856\div4$ c) $765\div5$ d) $984\div8$

Hints

- Break each dividend into familiar multiples of the divisor. - Make sure the parts add to the original dividend. - Divide each part and add the partial quotients.

Solution

1. For part a, $600 \div 3 = 200$, $30 \div 3 = 10$, and $9 \div 3 = 3$. Then $200 + 10 + 3 = 213$. 2. For part b, $800 \div 4 = 200$, $40 \div 4 = 10$, and $16 \div 4 = 4$. Then $200 + 10 + 4 = 214$. 3. For part c, $500 \div 5 = 100$, $250 \div 5 = 50$, and $15 \div 5 = 3$. Then $100 + 50 + 3 = 153$. 4. For part d, $800 \div 8 = 100$, $160 \div 8 = 20$, and $24 \div 8 = 3$. Then $100 + 20 + 3 = 123$.

Answer

a) $213$ b) $214$ c) $153$ d) $123$

Division by a one-digit number: place-value methods 51

Find each quotient. a) $520\div4$ b) $720\div6$ c) $850\div5$ d) $910\div7$

Hints

- Find a large familiar multiple of the divisor. - Divide the remaining part by the same divisor. - Add the partial quotients.

Solution

1. For part a, $520 = 400 + 120$. Then $400 \div 4 = 100$ and $120 \div 4 = 30$, so the quotient is $130$. 2. For part b, $720 = 600 + 120$. Then $600 \div 6 = 100$ and $120 \div 6 = 20$, so the quotient is $120$. 3. For part c, $850 = 500 + 350$. Then $500 \div 5 = 100$ and $350 \div 5 = 70$, so the quotient is $170$. 4. For part d, $910 = 700 + 210$. Then $700 \div 7 = 100$ and $210 \div 7 = 30$, so the quotient is $130$.

Answer

a) $130$ b) $120$ c) $170$ d) $130$

Division by a one-digit number: place-value methods 52

Find each quotient. a) $135\div5$ b) $168\div6$ c) $252\div4$ d) $344\div8$ e) $455\div7$

Hints

- Choose a large familiar multiple of the divisor below each dividend. - Divide that part and the remaining part. - Add the partial quotients.

Solution

1. For part a, $100 \div 5 = 20$ and $35 \div 5 = 7$. Then $20 + 7 = 27$. 2. For part b, $120 \div 6 = 20$ and $48 \div 6 = 8$. Then $20 + 8 = 28$. 3. For part c, $240 \div 4 = 60$ and $12 \div 4 = 3$. Then $60 + 3 = 63$. 4. For part d, $320 \div 8 = 40$ and $24 \div 8 = 3$. Then $40 + 3 = 43$. 5. For part e, $420 \div 7 = 60$ and $35 \div 7 = 5$. Then $60 + 5 = 65$.

Answer

a) $27$ b) $28$ c) $63$ d) $43$ e) $65$

Division by a one-digit number: place-value methods 53

Paul has $840$ stickers and shares them equally among $4$ friends. How many stickers does each friend receive?

Hints

- Break $840$ into two numbers that are easy to divide by $4$. - Divide each part by $4$. - Add the partial quotients.

Solution

1. Divide the total number of stickers by the number of friends: $840 \div 4$. 2. Break $840$ into compatible numbers: $840 = 800 + 40$. 3. Divide each part: $800 \div 4 = 200$ and $40 \div 4 = 10$. 4. Add the partial quotients: $200 + 10 = 210$.

Answer

Each friend receives $210$ stickers.

Division by a one-digit number: place-value methods 54

Three children buy a board game for $\$7.50$. They split the cost equally. How much does each child pay?

Hints

- Work in cents to divide a whole number. - Split the total number of cents into two numbers that are easy to divide by $3$. - Change your result back into dollars and cents.

Solution

1. Change the total price to cents: $\$7.50=750$ cents. 2. Divide the cents equally among the three children: $750\div3$. 3. Split the dividend: $600\div3=200$ and $150\div3=50$. Together, each child pays $200+50=250$ cents. 4. $250$ cents is $\$2.50$.

Answer

$\$2.50$ per child.

Division by a one-digit number: place-value methods 55

A bakery made $780$ rolls and divides them equally among $6$ baskets. How many rolls are in each basket?

Hints

- Break $780$ into two multiples of $6$. - Divide each part by $6$. - Add the partial quotients.

Solution

1. Divide the total number of rolls by the number of baskets: $780 \div 6$. 2. Break $780$ into compatible numbers: $780 = 600 + 180$. 3. Divide each part: $600 \div 6 = 100$ and $180 \div 6 = 30$. 4. Add the partial quotients: $100 + 30 = 130$.

Answer

Each basket contains $130$ rolls.

Division by a one-digit number: place-value methods 56

A student says, “The dividend $708$ has $0$ in the tens place, so $708\div3$ must have $0$ in the tens place too.” Is the student correct? Divide using place value and explain.

Hints

- When you cannot share all the hundreds equally, what can you do with any leftovers? - Remember that one hundred can be regrouped without changing its value. - Check the student's claim by multiplying your quotient by $3$.

Solution

1. Share $7$ hundreds among $3$ groups: each gets $2$ hundreds, with $1$ hundred left. 2. Trade the leftover hundred for $10$ tens. Each group gets $3$ tens, with $1$ ten left. 3. Trade the leftover ten for $10$ ones and add the $8$ ones. Then $18\div3=6$ ones per group. 4. So $708\div3=236$. The quotient has $3$ tens, not $0$.

Answer

No. $708\div3=236$. Regrouping a leftover hundred creates tens in the quotient.

Division by a one-digit number: place-value methods 57

Divide $732\div6$ using place value. After giving each group one hundred, what amount remains to divide? Finish and find the quotient.

Hints

- Share the hundreds, but keep track of what cannot yet be shared equally. - Trade leftover hundreds for tens before dividing the tens column. - Repeat the idea with any leftover tens before dividing ones.

Solution

1. Give each of the $6$ groups $1$ hundred. That uses $600$ of the $732$, leaving $132$: $1$ hundred, $3$ tens, and $2$ ones. 2. Trade the remaining hundred for $10$ tens. Divide the $13$ tens by $6$: each group gets $2$ tens, with $1$ ten left. 3. Trade the ten for $10$ ones and include $2$ more ones. Divide $12$ ones by $6$: each gets $2$ ones. 4. The quotient is $122$.

Answer

$132$ remains to divide. $732\div6=122$.

Division by a one-digit number: place-value methods 58

Each value in the first row is twice a number. Find half of that number and complete the table. <table> <tr> <td>Twice the number</td> <td>$480$</td> <td>$760$</td> <td>$1200$</td> <td>$3400$</td> <td>$6800$</td> </tr> <tr> <td>Half of the number</td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> </table>

Hints

- First find the number when you know twice that number. - After you find the number, how do you find half of it? - How many times as large is twice a number as half of the same number? - Can you find the target value in one step by dividing by a single number?

Solution

To move from twice a number to half of that number, divide by $2$ and then divide by $2$ again. This is the same as dividing by $4$. 1. $480 \div 2 = 240$, and $240 \div 2 = 120$. 2. $760 \div 2 = 380$, and $380 \div 2 = 190$. 3. $1200 \div 2 = 600$, and $600 \div 2 = 300$. 4. $3400 \div 2 = 1700$, and $1700 \div 2 = 850$. 5. $6800 \div 2 = 3400$, and $3400 \div 2 = 1700$.

Answer

Half of each original number is $120$, $190$, $300$, $850$, and $1700$.

Division by a one-digit number: place-value methods 59

Find each quotient. Use the first three equations to help with the last three. a) $800 \div 8 = \square$ b) $80 \div 8 = \square$ c) $16 \div 8 = \square$ d) $880 \div 8 = \square$ e) $896 \div 8 = \square$ f) $816 \div 8 = \square$

Hints

- Look for the parts $800$, $80$, and $16$ inside the larger dividends. - Divide each part by $8$, and then add the partial quotients. - Which of the first three dividends can be combined to make $896$?

Solution

1. Find the first three quotients: $800 \div 8 = 100$, $80 \div 8 = 10$, and $16 \div 8 = 2$. 2. For d), decompose $880$ as $800 + 80$: $100 + 10 = 110$. 3. For e), decompose $896$ as $800 + 80 + 16$: $100 + 10 + 2 = 112$. 4. For f), decompose $816$ as $800 + 16$: $100 + 2 = 102$.

Answer

a) $100$ b) $10$ c) $2$ d) $110$ e) $112$ f) $102$

Division by a one-digit number: place-value methods 60

Find each quotient. a) $321\div3$ b) $642\div6$ c) $436\div4$

Hints

- Split each dividend into a large multiple of $100$ and a smaller remainder. - Make sure both parts are divisible by the divisor. - Add the two partial quotients.

Solution

1. For a), decompose $321$ as $300 + 21$. Then $300 \div 3 = 100$ and $21 \div 3 = 7$, so the quotient is $107$. 2. For b), decompose $642$ as $600 + 42$. Then $600 \div 6 = 100$ and $42 \div 6 = 7$, so the quotient is $107$. 3. For c), decompose $436$ as $400 + 36$. Then $400 \div 4 = 100$ and $36 \div 4 = 9$, so the quotient is $109$.

Answer

a) $107$ b) $107$ c) $109$

Division by a one-digit number: place-value methods 61

Fill in the missing numbers in each decomposition, and then find the final quotient. a) $(\square \div 5) = (500 \div 5) + (45 \div 5) = 100 + 9 = 109$ b) $(\square \div 7) = (700 \div 7) + (70 \div 7) + (7 \div 7) = \square + \square + \square = \square$

Hints

- Each division has been broken into easier partial divisions. - Add the partial dividends to find the missing dividend. - For b), find each partial quotient before adding them.

Solution

1. For a), add the partial dividends: $500 + 45 = 545$. Therefore, $545 \div 5 = 109$. 2. For b), add the partial dividends: $700 + 70 + 7 = 777$. 3. Find the partial quotients: $700 \div 7 = 100$, $70 \div 7 = 10$, and $7 \div 7 = 1$. 4. Add them: $100 + 10 + 1 = 111$. Therefore, $777 \div 7 = 111$.

Answer

a) $545 \div 5 = 109$ b) $777 \div 7 = 100 + 10 + 1 = 111$

Division by a one-digit number: place-value methods 62

Find $456 \div 4$ by breaking $456$ into two parts that are easy to divide by $4$.

Hints

- Break $456$ into two numbers that are both multiples of $4$. - A multiple of $100$ may be a useful first part. - Add the partial quotients at the end.

Solution

1. Decompose $456$ as $400 + 56$, because both parts are divisible by $4$. 2. Divide the first part: $400 \div 4 = 100$. 3. Divide the second part: $56 \div 4 = 14$. 4. Add the partial quotients: $100 + 14 = 114$.

Answer

$114$

Division by a one-digit number: place-value methods 63

Complete the sequence of related division equations. a) $6000\div6=\square$ b) $420\div6=\square$ c) $6420\div6=\square$ d) $18\div6=\square$ e) $6438\div6=\square$

Hints

- Look for earlier dividends that can be combined to make a later dividend. - Divide convenient parts separately and add the partial quotients. - Reuse earlier results to solve the final division efficiently.

Solution

1. $6000 \div 6 = 1000$. 2. $420 \div 6 = 70$. 3. Since $6420 = 6000 + 420$, $6420 \div 6 = 1000 + 70 = 1070$. 4. $18 \div 6 = 3$. 5. Since $6438 = 6420 + 18$, $6438 \div 6 = 1070 + 3 = 1073$.

Answer

a) $1000$ b) $70$ c) $1070$ d) $3$ e) $1073$

Division by a one-digit number: place-value methods 64

Use the relationships among the equations to find the missing values. a) $2000 \div 2 = \square$ and $18 \div 2 = \square$. What is $2018 \div 2$? b) $9000 \div 9 = 1000$ and $630 \div 9 = 70$. Which dividend was formed by combining these parts? Complete $\square \div 9 = 1070$. c) $500 \div 5 = 100$ and $45 \div 5 = 9$. Find $545 \div 5$.

Hints

- Notice how each large dividend is built from easier parts. - Divide the parts separately, and then add the partial quotients. - How do the first two equations in each part combine to form the final equation?

Solution

1. For a), $2000 \div 2 = 1000$ and $18 \div 2 = 9$. Add the partial quotients: $1000 + 9 = 1009$. 2. For b), combine the partial dividends: $9000 + 630 = 9630$. Therefore, $9630 \div 9 = 1070$. 3. For c), add the partial quotients: $100 + 9 = 109$. Therefore, $545 \div 5 = 109$.

Answer

a) $2000 \div 2 = 1000$, $18 \div 2 = 9$, and $2018 \div 2 = 1009$ b) $9630 \div 9 = 1070$ c) $545 \div 5 = 109$

Division by a one-digit number: place-value methods 65

Find each quotient by breaking the dividend into two convenient parts that are both divisible by the divisor. Example: $1208 \div 4 = 1200 \div 4 + 8 \div 4 = 300 + 2 = 302$ a) $4212 \div 6$ b) $6456 \div 8$ c) $2135 \div 7$

Hints

- Look for a large part related to a basic multiplication fact and place value. - Make sure the remaining part is also divisible by the divisor. - Add the two partial quotients.

Solution

1. For a), decompose $4212$ as $4200 + 12$. Then $4200 \div 6 = 700$ and $12 \div 6 = 2$, so the quotient is $702$. 2. For b), decompose $6456$ as $6400 + 56$. Then $6400 \div 8 = 800$ and $56 \div 8 = 7$, so the quotient is $807$. 3. For c), decompose $2135$ as $2100 + 35$. Then $2100 \div 7 = 300$ and $35 \div 7 = 5$, so the quotient is $305$.

Answer

a) $702$ b) $807$ c) $305$

Division by a one-digit number: place-value methods 66

Bima wants to find $3618 \div 6$. Bima considers breaking $3618$ into either $3000 + 618$ or $3600 + 18$. a) Which decomposition makes the quotient easier to find? Explain briefly. b) Use the easier decomposition to find $3618 \div 6$. c) Use a similarly efficient decomposition to find $4540 \div 5$.

Hints

- Which proposed parts are easy multiples of $6$? - Use related basic multiplication facts to identify convenient dividends. - For c), look for a nearby multiple of $5$ that is easy to divide.

Solution

1. For a), $3600 + 18$ is more efficient because both parts are immediately recognizable multiples of $6$. 2. For b), $3600 \div 6 = 600$ and $18 \div 6 = 3$. Therefore, $3618 \div 6 = 603$. 3. For c), decompose $4540$ as $4500 + 40$. Then $4500 \div 5 = 900$ and $40 \div 5 = 8$, so the quotient is $908$.

Answer

a) $3600 + 18$, because both parts are easy to divide by $6$ b) $603$ c) $908$

Division by a one-digit number: place-value methods 67

Look at the three division problems. Solve the easiest one first, and then use its quotient to find the other two mentally. a) $1842 \div 6$ b) $1800 \div 6$ c) $1794 \div 6$

Hints

- Which dividend is easiest to divide by $6$? - How much greater or less are the other dividends? - How does the quotient change when the dividend changes by $6$?

Solution

1. The easiest problem is $1800 \div 6 = 300$. 2. The dividend $1842$ is $42$ greater than $1800$. Since $42 \div 6 = 7$, add $7$: $300 + 7 = 307$. 3. The dividend $1794$ is $6$ less than $1800$. Since $6 \div 6 = 1$, subtract $1$: $300 - 1 = 299$.

Answer

a) $307$ b) $300$ c) $299$

Division by a one-digit number: place-value methods 68

Fill in the missing numbers to solve each division problem by decomposition. a) $2832 \div 4 = (2800 \div 4) + (\square \div 4) = 700 + \square = \square$ b) $8127 \div 9 = (8100 \div 9) + (\square \div 9) = \square + \square = \square$

Hints

- Find the missing part that combines with the given part to make the original dividend. - Divide both parts separately. - Add the partial quotients.

Solution

1. For a), decompose $2832$ as $2800 + 32$. Then $2800 \div 4 = 700$ and $32 \div 4 = 8$, so the quotient is $708$. 2. For b), decompose $8127$ as $8100 + 27$. Then $8100 \div 9 = 900$ and $27 \div 9 = 3$, so the quotient is $903$.

Answer

a) $2832 \div 4 = (2800 \div 4) + (32 \div 4) = 700 + 8 = 708$ b) $8127 \div 9 = (8100 \div 9) + (27 \div 9) = 900 + 3 = 903$

Division by a one-digit number: place-value methods 69

Which equation must be incorrect? Justify your choice with an estimate. a) $3456 \div 9=384$ b) $1944 \div 6=324$ c) $5565 \div 7=95$

Hints

- Estimate each quotient using a nearby compatible number. - Compare the number of digits in each estimate with the stated quotient. - Look for the result that differs greatly from its estimate.

Solution

1. For a), $3456 \div 9\approx3600 \div 9=400$. The stated quotient $384$ is reasonable. 2. For b), $1944 \div 6\approx1800 \div 6=300$. The stated quotient $324$ is reasonable. 3. For c), $5565 \div 7\approx5600 \div 7=800$. The stated quotient $95$ is far too small. The exact quotient is $795$.

Answer

Equation c) is incorrect. Since $5565 \div 7\approx800$, the quotient must be in the hundreds, not $95$.

Division by a one-digit number: place-value methods 70

Estimate each quotient first. Then calculate the exact quotient. a) $66 \div 6$ b) $606 \div 6$ c) $6006 \div 6$

Hints

- Choose a nearby compatible number divisible by $6$. - Watch for zeros that must be written in the quotient. - Compare each exact quotient with its estimate.

Solution

1. For a), $66 \div 6\approx60 \div 6=10$. The exact quotient is $11$. 2. For b), $606 \div 6\approx600 \div 6=100$. The exact quotient is $101$. 3. For c), $6006 \div 6\approx6000 \div 6=1000$. The exact quotient is $1001$.

Answer

a) Estimate: $10$; exact quotient: $11$ b) Estimate: $100$; exact quotient: $101$ c) Estimate: $1000$; exact quotient: $1001$

Division by a one-digit number: place-value methods 71

Find $132 \div 6$. Then find $176 \div 8$. Compare the two quotients. What do you notice?

Hints

- Solve both division problems separately. - Break each dividend into familiar multiples of the divisor. - Compare the two quotients.

Solution

1. Break $132$ into $120 + 12$. Then $120 \div 6 = 20$ and $12 \div 6 = 2$, so $132 \div 6 = 22$. 2. Break $176$ into $160 + 16$. Then $160 \div 8 = 20$ and $16 \div 8 = 2$, so $176 \div 8 = 22$. 3. The quotients are equal.

Answer

$132 \div 6 = 22$ and $176 \div 8 = 22$. The quotients are equal.

Division by a one-digit number: place-value methods 72

How many groups of $8$ are in $584$? Use partial quotients by breaking $584$ into two numbers that are easy to divide by $8$.

Hints

- Find a large multiple of $8$ close to $584$. - Divide that part and the remaining part by $8$. - Add the partial quotients.

Solution

1. Write the division problem $584 \div 8$. 2. Break $584$ into $560 + 24$. 3. Divide each part: $560 \div 8 = 70$ and $24 \div 8 = 3$. 4. Add the partial quotients: $70 + 3 = 73$.

Answer

There are $73$ groups of $8$ in $584$.

Division by a one-digit number: place-value methods 73

A plant nursery received $168$ tulip bulbs. a) The nursery divides the bulbs equally among $6$ planter boxes. How many bulbs go in each box? b) How many bulbs would go in each box if the nursery used $8$ planter boxes instead?

Hints

- For part a, break $168$ into numbers that are easy to divide by $6$. - For part b, find a different way to break apart $168$ so both parts are easy to divide by $8$. - When the same number of bulbs is divided among more boxes, should each box get more or fewer bulbs?

Solution

1. For part a, break $168$ into compatible numbers: $168 = 120 + 48$. 2. Divide each part by $6$: $120 \div 6 = 20$ and $48 \div 6 = 8$. Then add: $20 + 8 = 28$. 3. For part b, use a different decomposition: $168 = 160 + 8$. 4. Divide each part by $8$: $160 \div 8 = 20$ and $8 \div 8 = 1$. Then add: $20 + 1 = 21$.

Answer

a) Each of the $6$ boxes gets $28$ tulip bulbs. b) Each of the $8$ boxes would get $21$ tulip bulbs.

Division by a one-digit number: place-value methods 74

Fill in the missing numbers in each partial-quotients method. a) $84 \div 4$ $80 \div 4 = \square$ $4 \div 4 = \square$ so $84 \div 4 = \square$ b) $96 \div 3$ $\square \div 3 = 30$ $6 \div 3 = \square$ so $96 \div 3 = \square$ c) $75 \div 5$ $50 \div 5 = \square$ $25 \div 5 = \square$ so $75 \div 5 = \square$

Hints

- Identify the two parts that make the original dividend. - Use multiplication to work backward when a partial dividend is missing. - Add the partial quotients.

Solution

1. For part a, $80 \div 4 = 20$ and $4 \div 4 = 1$. Then $20 + 1 = 21$. 2. For part b, the missing partial dividend is $90$, because $90 \div 3 = 30$. Also, $6 \div 3 = 2$. Then $30 + 2 = 32$. 3. For part c, $50 \div 5 = 10$ and $25 \div 5 = 5$. Then $10 + 5 = 15$.

Answer

a) $80 \div 4 = 20$, $4 \div 4 = 1$, so $84 \div 4 = 21$ b) $90 \div 3 = 30$, $6 \div 3 = 2$, so $96 \div 3 = 32$ c) $50 \div 5 = 10$, $25 \div 5 = 5$, so $75 \div 5 = 15$

Division by a one-digit number: place-value methods 75

Find each quotient using partial quotients. Break the dividend so that the first part is ten times the divisor. a) $52 \div 4$ b) $78 \div 6$ c) $91 \div 7$ d) $104 \div 8$ What do you notice about the quotients?

Hints

- Write two partial-division steps for each problem. - Make the first partial dividend ten times the divisor. - Divide the remaining part by the same divisor. - Add the partial quotients and compare the results.

Solution

1. For part a, $40 \div 4 = 10$ and $12 \div 4 = 3$. Then $10 + 3 = 13$. 2. For part b, $60 \div 6 = 10$ and $18 \div 6 = 3$. Then $10 + 3 = 13$. 3. For part c, $70 \div 7 = 10$ and $21 \div 7 = 3$. Then $10 + 3 = 13$. 4. For part d, $80 \div 8 = 10$ and $24 \div 8 = 3$. Then $10 + 3 = 13$. 5. All four quotients are equal.

Answer

a) $13$ b) $13$ c) $13$ d) $13$ All four quotients are equal.

Division by a one-digit number: place-value methods 76

Use partial quotients to determine how many times as great the second number is as the first. a) $3$ and $114$ b) $9$ and $153$

Hints

- Divide the second number by the first number. - Decompose the dividend into parts that are easy to divide. - For $114 \div 3$, begin with a large multiple of $3$ that you know.

Solution

1. For a), compute $114 \div 3$. Decompose $114$ as $90 + 24$: $90 \div 3 = 30$ and $24 \div 3 = 8$. Then $30 + 8 = 38$. 2. For b), compute $153 \div 9$. Decompose $153$ as $90 + 63$: $90 \div 9 = 10$ and $63 \div 9 = 7$. Then $10 + 7 = 17$.

Answer

a) $114$ is $38$ times as great as $3$. b) $153$ is $17$ times as great as $9$.

Division by a one-digit number: place-value methods 77

A school event has $135$ muffins. Each serving tray holds exactly $9$ muffins. A staff member says, “We need at least $20$ trays to display all the muffins at once.” Is the staff member correct? Justify your answer with a calculation.

Hints

- First calculate the exact number of trays needed. - Break $135$ into numbers that are easy to divide by $9$. - Compare your result with $20$.

Solution

1. Find the number of trays needed: $135 \div 9$. 2. Break $135$ into compatible numbers: $135 = 90 + 45$. 3. Divide each part: $90 \div 9 = 10$ and $45 \div 9 = 5$. Then add: $10 + 5 = 15$. 4. Only $15$ trays are needed, so the statement that at least $20$ are needed is false.

Answer

No. Only $15$ trays are needed because $135 \div 9 = 15$.

Division by a one-digit number: place-value methods 78

For a project week, $192$ small clay pots are divided equally among $6$ worktables. How many pots are placed on each table? At one table, $4$ children divide that table’s pots equally. How many pots does each child get?

Hints

- Break $192$ into two multiples of $6$. - First find the number of pots on one table. - Then divide that amount equally among $4$ children.

Solution

1. Divide the pots among the tables: $192 \div 6 = (180 \div 6) + (12 \div 6) = 30 + 2 = 32$. 2. Divide one table’s $32$ pots among $4$ children: $32 \div 4 = 8$.

Answer

Each table has $32$ pots. Each of the $4$ children gets $8$ pots.

Division by a one-digit number: place-value methods 79

Elena, a gardener, divides $855$ seedlings equally among $9$ garden beds. a) How many seedlings go in each bed? b) Tomás breaks $855$ into $810 + 45$. Explain why this is an efficient choice for mental division or partial quotients.

Hints

- Look for a large multiple of $9$ close to $855$. - Think about how a useful decomposition can turn one difficult division into easier familiar facts. - Explain why each chosen part works cleanly with the divisor.

Solution

1. Divide each part by $9$: $810 \div 9 = 90$ and $45 \div 9 = 5$. 2. Add the partial quotients: $90 + 5 = 95$. Each bed gets $95$ seedlings. 3. The decomposition is efficient because both $810$ and $45$ are familiar multiples of $9$ and divide evenly by $9$.

Answer

a) Each bed gets $95$ seedlings. b) The decomposition is efficient because both $810$ and $45$ are familiar multiples of $9$.

Division by a one-digit number: place-value methods 80

A printing company packs $861$ posters equally into $7$ shipping boxes. How many posters go in each box?

Hints

- Start with a large familiar multiple of $7$. - Break the remaining amount into another multiple of $7$. - Add the partial quotients.

Solution

1. Break $861$ into $700 + 140 + 21$, all multiples of $7$. 2. Divide each part: $700 \div 7 = 100$, $140 \div 7 = 20$, and $21 \div 7 = 3$. 3. Add the partial quotients: $100 + 20 + 3 = 123$.

Answer

Each box contains $123$ posters.

Division by a one-digit number: place-value methods 81

Use related facts to solve. a) Evaluate $450 \div 5$. Which related fact helps? b) Without calculating exactly, is $450 \div 9$ greater than or less than $450 \div 5$? Explain. c) Evaluate $450 \div 9$.

Hints

- Use related facts with $45$. - For the same dividend, compare how the divisor affects the quotient. - Check each quotient with multiplication.

Solution

1. a) The related fact $45 \div 5 = 9$ gives $450 \div 5 = 90$. 2. b) With the same dividend, dividing by the greater divisor gives the smaller quotient, so $450 \div 9 < 450 \div 5$. 3. c) The related fact $45 \div 9 = 5$ gives $450 \div 9 = 50$.

Answer

a) $90$; related fact: $45 \div 5 = 9$ b) Less than, because the divisor is greater. c) $50$

Division by a one-digit number: place-value methods 82

A farmer collected $840$ eggs and packs $6$ eggs in each carton. a) How many cartons can the farmer fill? b) If the farmer used cartons that held only $3$ eggs each, would more or fewer cartons be needed? Explain.

Hints

- Break $840$ into numbers that are easy to divide by $6$. - Add the partial quotients for part a. - For part b, compare the two carton sizes without finding the exact second quotient.

Solution

1. For part a, divide $840$ by $6$. Break apart $840 = 600 + 240$. 2. Divide each part: $600 \div 6 = 100$ and $240 \div 6 = 40$. Then add: $100 + 40 = 140$. 3. For part b, cartons holding $3$ eggs are half the size of cartons holding $6$. Smaller groups mean more groups are needed for the same total.

Answer

a) The farmer can fill $140$ cartons. b) More cartons would be needed because each carton holds fewer eggs.

Division by a one-digit number: place-value methods 83

A school event receives $672$ juice bottles packed in cases of $6$ bottles each. How many cases are delivered?

Hints

- Break $672$ into familiar multiples of $6$. - Divide each part by $6$. - Add the partial quotients.

Solution

1. Break $672$ into $600 + 60 + 12$, all divisible by $6$. 2. Divide each part: $600 \div 6 = 100$, $60 \div 6 = 10$, and $12 \div 6 = 2$. 3. Add the partial quotients: $100 + 10 + 2 = 112$.

Answer

A total of $112$ cases are delivered.

Division by a one-digit number: place-value methods 84

Calculate each pair and compare the quotients. a) $360 \div 3$ and $306 \div 3$ b) $840 \div 4$ and $804 \div 4$ c) $550 \div 5$ and $505 \div 5$ What digit pattern do you notice in each pair of quotients?

Hints

- Calculate each quotient first. - Compare the place of the zero in each pair. - Identify which digits remain the same.

Solution

1. a) $360 \div 3 = 120$ and $306 \div 3 = 102$. 2. b) $840 \div 4 = 210$ and $804 \div 4 = 201$. 3. c) $550 \div 5 = 110$ and $505 \div 5 = 101$. 4. In each pair, the same nonzero digits appear, while the zero shifts from the ones place to the tens place in the quotient.

Answer

a) $120$ and $102$ b) $210$ and $201$ c) $110$ and $101$ In each pair, the zero and the final nonzero digit switch places in the quotient.

Division by a one-digit number: place-value methods 85

You know that $900 \div 3 = 300$. Use this fact to find each new quotient without starting over. a) $900 \div 6$ b) $450 \div 3$ Explain how each result follows from the given equation.

Hints

- Compare each new divisor or dividend with the given equation. - Think about sharing the same amount among twice as many groups. - Think about sharing half as much among the same number of groups.

Solution

1. In a), the dividend stays fixed and the divisor doubles from $3$ to $6$, so the quotient is halved: $300 \div 2 = 150$. 2. In b), the divisor stays fixed and the dividend is halved from $900$ to $450$, so the quotient is also halved: $300 \div 2 = 150$.

Answer

a) $150$ b) $150$ Doubling the divisor halves the quotient in a); halving the dividend halves the quotient in b).

Division by a one-digit number: place-value methods 86

Find the four quotients. Then order the division expressions from the smallest quotient to the greatest quotient. a) $630 \div 3$ b) $360 \div 3$ c) $603 \div 3$ d) $306 \div 3$

Hints

- Find each quotient separately. - Use place-value parts to divide by $3$. - Compare the hundreds, tens, and ones in the quotients.

Solution

1. The quotients are $630 \div 3 = 210$, $360 \div 3 = 120$, $603 \div 3 = 201$, and $306 \div 3 = 102$. 2. In increasing order, $102 < 120 < 201 < 210$, so the order is d, b, c, a.

Answer

a) $630 \div 3 = 210$ b) $360 \div 3 = 120$ c) $603 \div 3 = 201$ d) $306 \div 3 = 102$ From smallest quotient to greatest quotient: d), b), c), a).

Division by a one-digit number: place-value methods 87

At a field day, $168$ students are first divided into teams of $6$. How many teams are formed? Then calculate how many teams would be formed if each team had $4$ students instead. Would the smaller team size create more or fewer teams?

Hints

- Divide $168$ by each team size. - Use a decomposition that works well with each divisor. - Compare the two quotients.

Solution

1. For teams of $6$, break $168$ into compatible numbers: $168 = 120 + 48$. Then $120 \div 6 = 20$ and $48 \div 6 = 8$, so $20 + 8 = 28$ teams. 2. For teams of $4$, use $168 = 160 + 8$. Then $160 \div 4 = 40$ and $8 \div 4 = 2$, so $40 + 2 = 42$ teams. 3. Since $42 > 28$, the smaller team size creates more teams.

Answer

Teams of $6$ produce $28$ teams. Teams of $4$ produce $42$ teams, so the smaller team size creates more teams.

Division by a one-digit number: place-value methods 88

Find $744 \div 8$ using partial quotients. Show two different ways to break $744$ into two or three numbers that are easy to divide by $8$.

Hints

- Find a large multiple of $8$ close to $744$. - Choose a different large multiple of $8$ for the second method. - Make sure every part is divisible by $8$.

Solution

1. One decomposition is $744 = 720 + 24$. Then $720 \div 8 = 90$ and $24 \div 8 = 3$, so the quotient is $93$. 2. Another decomposition is $744 = 640 + 104$. Then $640 \div 8 = 80$ and $104 \div 8 = 13$. The second partial quotient can be found from $104 = 80 + 24$: $80 \div 8 = 10$ and $24 \div 8 = 3$. Thus $80 + 13 = 93$.

Answer

The quotient is $93$. One method uses $720 \div 8 = 90$ and $24 \div 8 = 3$. Another method uses $640 \div 8 = 80$ and $104 \div 8 = 13$.

Division by a one-digit number: place-value methods 89

Four children have saved $\$468$. They divide the money equally. How much money does each child receive?

Hints

- Break $468$ into parts that are easy to divide by $4$. - Divide each part by $4$. - Add the partial quotients.

Solution

1. Break $468$ into $400 + 60 + 8$, with each part divisible by $4$. 2. Divide each part: $400 \div 4 = 100$, $60 \div 4 = 15$, and $8 \div 4 = 2$. 3. Add the partial quotients: $100 + 15 + 2 = 117$.

Answer

Each child receives $\$117$.

Division by a one-digit number: place-value methods 90

Compare each pair of quotients. Write $<$, $>$, or $=$ in the box. Briefly justify your answer to part c). a) $400 \div 5 \quad \square \quad 400 \div 8$ b) $240 \div 4 \quad \square \quad 480 \div 4$ c) $180 \div 2 \quad \square \quad 360 \div 4$

Hints

- Look for what stays the same in each pair. - With the same dividend, think about how the divisor affects the quotient. - With the same divisor, think about how the dividend affects the quotient. - In part c), compare how both numbers change.

Solution

1. In part a), the dividend is the same. Dividing by the smaller divisor gives the greater quotient: $400 \div 5 = 80$ and $400 \div 8 = 50$, so $80 > 50$. 2. In part b), the divisor is the same. The greater dividend gives the greater quotient: $240 \div 4 = 60$ and $480 \div 4 = 120$, so $60 < 120$. 3. In part c), both the dividend and divisor are doubled: $180 \times 2 = 360$ and $2 \times 2 = 4$. The quotient stays the same, and $180 \div 2 = 360 \div 4 = 90$.

Answer

a) $>$ b) $<$ c) $=$; both the dividend and divisor are doubled, so the quotient stays the same.

Division by a one-digit number: place-value methods 91

Lukas breaks $876$ into $600 + 240 + 36$ to find $876 \div 6$. a) Use Lukas’s decomposition to find the quotient. b) Find a different way to break $876$ into two or three parts that are easy to divide by $6$. Show your work.

Hints

- Look for familiar multiples of $6$ or $60$. - Make sure the parts add to $876$. - Divide each part and add the partial quotients.

Solution

1. Using Lukas’s method, $600 \div 6 = 100$, $240 \div 6 = 40$, and $36 \div 6 = 6$. Then $100 + 40 + 6 = 146$. 2. One alternative is $876 = 840 + 36$. Then $840 \div 6 = 140$ and $36 \div 6 = 6$, so $140 + 6 = 146$.

Answer

a) $146$ b) One example is $840 \div 6 = 140$ and $36 \div 6 = 6$, so the quotient is $146$.

Division by a one-digit number: place-value methods 92

A gardener has $156$ flower bulbs and plants exactly $6$ bulbs in each row. How many rows can the gardener plant? Show how you break apart the division problem.

Hints

- Break $156$ into two multiples of $6$. - Divide each part by $6$. - Add the partial quotients.

Solution

1. The division problem is $156 \div 6$. 2. Break $156$ into $120 + 36$. 3. Divide each part: $120 \div 6 = 20$ and $36 \div 6 = 6$. 4. Add the partial quotients: $20 + 6 = 26$.

Answer

The gardener can plant $26$ rows.

Division by a one-digit number: place-value methods 93

A field trip needs $336$ small juice bottles. The bottles come in cases of $8$. A teacher says, “Exactly $40$ cases will be enough.” Is the teacher correct? Justify your answer with a calculation.

Hints

- First calculate the exact number of cases needed. - Break $336$ into multiples of $8$. - Compare the quotient with $40$.

Solution

1. Find the number of cases needed: $336 \div 8$. 2. Break $336$ into compatible numbers: $336 = 320 + 16$. 3. Divide each part: $320 \div 8 = 40$ and $16 \div 8 = 2$. 4. Add: $40 + 2 = 42$. Since $42 \ne 40$, the teacher is not correct.

Answer

No. The trip needs $42$ cases, so $40$ cases are not enough.

Division by a one-digit number: place-value methods 94

Maya claims, “When I divide a four-digit number by $6$, the quotient always has $3$ digits.” a) Give an example for which Maya’s claim is true. b) Give a counterexample for which the quotient has $4$ digits. c) Complete the rule: The quotient of a four-digit number divided by $6$ has $4$ digits whenever the thousands digit is ...

Hints

- Compare the thousands digit with the divisor $6$. - Decide whether division begins with the thousands digit or with the first two digits. - Choose one four-digit dividend on each side of that threshold.

Solution

1. For a), choose a four-digit number with a thousands digit less than $6$, such as $1200$. Then $1200 \div 6=200$, which has $3$ digits. 2. For b), choose a four-digit number with a thousands digit at least $6$, such as $6000$. Then $6000 \div 6=1000$, which has $4$ digits. 3. The thousands digit must be at least $6$.

Answer

a) For example, $1200 \div 6=200$. b) For example, $6000 \div 6=1000$. c) ... at least $6$.

Division by a one-digit number: place-value methods 95

Two classes collect paper for a school recycling project. Ms. Green's class collected $840\,\text{lb}$ and divides it equally among $4$ containers. Mr. Lee's class collected $750\,\text{lb}$ and divides it equally among $3$ containers. Which class has more paper in each container?

Hints

- First find how many pounds of paper are in one container for each class. - Break each dividend into hundreds and tens that are easy to divide by the given divisor. - Compare the two amounts per container.

Solution

1. Find the amount in each container for Ms. Green's class: $840\,\text{lb} \div 4 = (800\,\text{lb} \div 4) + (40\,\text{lb} \div 4) = 200\,\text{lb} + 10\,\text{lb} = 210\,\text{lb}$. 2. Find the amount in each container for Mr. Lee's class: $750\,\text{lb} \div 3 = (600\,\text{lb} \div 3) + (150\,\text{lb} \div 3) = 200\,\text{lb} + 50\,\text{lb} = 250\,\text{lb}$. 3. Compare the amounts: $250\,\text{lb} > 210\,\text{lb}$.

Answer

Mr. Lee's class has more paper in each container, with $250\,\text{lb}$ compared with $210\,\text{lb}$ for Ms. Green's class.

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

Grade 4 math worksheets

Division by a one-digit number: place-value methods

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